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edited Jan 12 2011 at 22:40 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
This is a somewhat dumb attempt at weakening solidity. My feeling is that a useful weakening of solidity shouldn't be possible. (Certain weakenings exist, though, as pointed out below by Owen.) Is there an intelligent way to weaken the notion of solidity for von Neumann algebras? (I concede that there really isn't an intelligent reason to try to!)
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edited Jan 12 2011 at 22:26 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
This is a somewhat dumb attempt at weakening solidity. My feeling is that a useful weakening of solidity shouldn't be possible. (Certain weakenings exist, though, as pointed out below by Owen.) Is there an intelligent way to weaken the notion of solidity for von Neumann algebras? (I concede that there really isn't an intelligent reason to try to!)
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edited Jan 12 2011 at 13:03 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
I think the answer may be no. What happens if you tensor a solid factor with a non-injective McDuff factor, for example $\mathcal{M} \otimes (\mathcal{M}\otimes \mathcal{R})$, with $\mathcal{M}$ solid? The resulting thing cannot be solid, since it has a noninjective subfactor with property $\Gamma$. Can this resulting thing be $\Gamma$-solid? (If this is so, then a $\Gamma$-solid factor can have property $\Gamma$!) We then may ask whether every $\Gamma$-solid factor has property $\Gamma$.
This is a somewhat dumb attempt at weakening solidity. My feeling is that a useful weakening of solidity shouldn't be possible. Is there an intelligent way to weaken the notion of solidity for von Neumann algebras? (I concede that there really isn't an intelligent reason to try to!)
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Post Undeleted by Anton Geraschenko♦♦
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occurred Jan 4 2011 at 0:58
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Post Deleted by Jon Bannon
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occurred Dec 11 2010 at 15:58
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edited Dec 7 2010 at 18:37 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
I think the answer may be no. What happens if you tensor a solid factor with a non-injective McDuff factor, for example $\mathcal{M} \otimes (\mathcal{M}\otimes \mathcal{R})$, with $\mathcal{M}$ solid? The resulting thing cannot be solid, since it has a noninjective subfactor with property $\Gamma$. Can this resulting thing be $\Gamma$-solid? (If this is so, then a $\Gamma$-solid factor can have property $\Gamma$!) We then may ask whether every $\Gamma$-solid factor has property $\Gamma$.
This is a somewhat dumb attempt at weakening solidity. My feeling is that a useful weakening of solidity shouldn't be possible. Is there an intelligent way to weaken the notion of solidity for von Neumann algebras? (I concede that there really isn't an intelligent reason to try to!)
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edited Dec 7 2010 at 18:28 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
I think the answer may be no. What happens if you tensor a solid factor with a non-injective McDuff factor? The resulting thing cannot be solid, since it has a noninjective subfactor with property $\Gamma$. Can this resulting thing be $\Gamma$-solid? (In that caseIf this is so, then a $\Gamma$-solid factor can have property $\Gamma$!)
This is a somewhat dumb attempt at weakening solidity. My feeling is that a useful weakening of solidity shouldn't be possible.
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edited Dec 7 2010 at 18:21 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
I think the answer may be no. What happens if you tensor a solid factor with a non-injective McDuff factor? The resulting thing cannot be solid, since it has a noninjective subfactor with property $\Gamma$. Can such a this resulting thing be $\Gamma$-solid? (In that case, a $\Gamma$-solid factor can have property $\Gamma$!)
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edited Dec 7 2010 at 18:16 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
I think the answer may be no. What happens if you tensor a solid factor with a non-injective McDuff factor? The resulting thing cannot be solid, since it has a noninjective subfactor with property $\Gamma$. Can such a thing be $\Gamma$-solid?
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edited Dec 7 2010 at 18:09 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
(I ask this question with full knowledge that there is no really compelling reason to answer it. I'm just curious if it is obvious that no weaker idea of solidity can be expected.)
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edited Dec 7 2010 at 17:14 |
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
(I ask this question with full knowledge that there is no really compelling reason to answer it. I'm just curious if it is obvious that no weaker idea of solidity can be expected.)
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asked Dec 7 2010 at 16:07 |
Weakly solid factors?
A type $II_{1}$ factor $\mathcal{M}$ is solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ is injective. (See Ozawa's excellent paper: http://arxiv.org/abs/math/0302082)
A type $II_{1}$ factor $M$ with trace $\tau$ has Property $\Gamma$ if for every finite subset ${ x_{1}, x_{2},..., x_{n} } \subseteq M$ and each $\epsilon >0$, there is a unitary element $u$ in $M$ with $\tau (u)=0$ and $||ux_{j}-x_{j}u||_{2}<\epsilon$ for all $1 \leq j \leq n$. (Here $||T||_2=(\tau(T^{*}T))^{1/2}$ for $T\in M$.)
We say that a type $II_{1}$ factor $M$ with trace $\tau$ is $\Gamma$-solid if for every diffuse von Neumann subalgebra $\mathcal{A}$ of $\mathcal{M}$, the relative commutant $\mathcal{A}'\cap \mathcal{M}$ has property $\Gamma$.
Since every injective von Neumann algebra has property $\Gamma$, the notion of $\Gamma$-solidity is weaker than that of solidity.
Is every $\Gamma$-solid factor solid?
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